JEE Main202329 Jan 2023Evening ShiftMathematicsDefinite IntegrationActual
The value of the integral ∫ 1 2 t 4 + 1 t 6 + 1 d t is :
Options
- Atan - 1 1 2 + 1 3 tan - 1 8 - π 3
- Btan - 1 2 - 1 3 tan - 1 8 + π 3
- Ctan - 1 2 + 1 3 tan - 1 8 - π 3
- Dtan - 1 1 2 - 1 3 tan - 1 8 + π 3
Correct answer
C. tan - 1 2 + 1 3 tan - 1 8 - π 3
Step-by-step solution
Given, ∫ 1 2 t 4 + 1 t 6 + 1 d t = ∫ 1 2 t 4 + 1 t 2 + 1 t 4 - t 2 + 1 d t ∵ a 3 + b 3 = ( a + b ) ( a 2 - a b + b 2 ) = ∫ 1 2 t 4 + 1 - t 2 + t 2 t 2 + 1 t 4 - t 2 + 1 d t = ∫ 1 2 t 4 + 1 - t 2 t 2 + 1 t 4 - t 2 + 1 + t 2 t 2 + 1 t 4 - t 2 + 1 d t = ∫ 1 2 1 t 2 + 1 + t 2 t 2 + 1 t 4 - t 2 + 1 d t = ∫ 1 2 1 t 2 + 1 + t 2 t 6 + 1 d t = ∫ 1 2 1 t 2 + 1 d t + ∫ 1 2 t 2 t 3 2 + 1 d t = tan - 1 t 1 2 + 1 3 ∫ 1 2 3 t 2 t 3 2 + 1 d t = tan - 1 2 - tan - 1 1 + 1 3